To solve this related rates (of change) problem:
Let #y# = the height of the balloon and let #theta# = the angle of elevation.
We are told that #(dy)/(dt)=8# ft/sec.
We are asked to find #(d theta)/(dt)# when #y=25# ft.
Draw a right triangle with base = 60 ft (that doesn't change), height #y# and angle opposite height #theta#.
Then # tan theta = y/60# and #y=60 tan theta#.
Differentiating with respect to #t# gives us:
#d/(dt)(y)=d/(dt)(60 tan theta)#.
#(dy)/(dt) = 60 sec^2 theta (d theta)/ (dt)#.
We are asked to find #(d theta)/(dt)# when #y=25#.
We have: #8= 60 sec^2 theta (d theta)/ (dt)#, so
#(d theta)/ (dt)=8/60 cos^2 theta = 2/15 cos^2 theta#.
We need #cos theta# when #y=25#.
With base = 60 and height = 25, we get hypotneuse #c= sqrt (60^2 + 25^2) = sqrt ((5*12)^2+(5*5)^2)=5sqrt ((12)^2+(5)^2) = 5*13 = 65#.
So, when #y=25#, we have: #cos theta = 60/65=12/13#.
So
#(d theta)/ (dt) = 2/15 cos^2 theta= 2/15 (12/13)^2 = 96/845# radians / sec
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(Remember, in order to use #d/(d theta)(tan theta) = sec^2 theta#, we must have #theta# either a real number or the radian (not degree) measure of an angle.)